eq16 := r(t) = d[vol]*V/(KUS*V^2+L*tau);
ex := InertForm:-MakeInert(convert(rhs(eq16), fullparfrac, KUS)):
eq17 := r(t) = op(1, ex)/expand(simplify(op(2, ex)));

restart:
with(Physics[Vectors]):
Setup(mathematicalnotation = true):
eq := r(t) = 2*t^2*_i+16*_j+(10*t-12)*_k;
v := diff(rhs(eq), t);
V := eval(v, t = 10);
simplify(Norm(V));

restart;
with(Physics[Vectors]);
Setup(mathematicalnotation = true);
eq := r(t) = 3*cos(5*t)*_i+sin(5*t)*_j+3*sin(5*t)*_k;
Norm(rhs(eq));#Calculate Norm
plots:-spacecurve([Component(rhs(eq), 1), Component(rhs(eq), 2), Component(rhs(eq), 3)], t = 0 .. 4*Pi, color = pink);

Have fun!

restart;
q1 := 9045.084972*(diff(z[1](t), t$2))+863728.7570*z[1](t) = -1963.525491562420*sin(20*t);
q2 := 3454.915028*(diff(z[2](t), t$2))+2.261271243*10^6*z[2](t) = -286.4745084375789*sin(20*t); icy := seq([z[i](0) = 0, (D(z[i]))(0) = 0], i = 1 .. 2, 1);

so := dsolve({q1, q2, seq(icy[i][], i = 1 .. 2), A(t) = diff(z[1](t), t$2), B(t) = diff(z[2](t), t$2)}, numeric); so(1);

plots:-odeplot(so, [[t, A(t)], [t, B(t)]], t = 0 .. 1, color = [red, blue], legend = ["diff(z[1](t), t$2)", "diff(z[2](t), t$2)"], legendstyle = [font = [times, bold, 20]]);

Have fun!

w := proc (x, y) options operator, arrow; piecewise(y <= .5, -2*tanh(y-.25), .5 < y, 2*tanh(.75-y)) end proc;
Matrix([seq([seq(w(x, y), x = 0 .. 10)], y = 0 .. 10)]);

#or;

Matrix([seq([seq(w(x, y), y = 0 .. 10)], x = 0 .. 10)]);

You don't gives values of constans,so I assuming.

restart;
Digits := 20;
Theta := (1/3)*Pi; Upsilon := 1/10;#assume!
eq := ((D@@2)(u))(r) = (-(D(u))(r)^2*u(r)+((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2)*((D(u))(r)*cot(Theta)+2*u(r)))/((D(u))(r)^2-((Upsilon-1)*(1/2))*(1-u(r)^2-(D(u))(r)^2));
sol := dsolve({eq, u(13.75) = .7787, (D(u))(13.75) = .344037}, numeric, abserr = 1.*10^(-16));
plots:-odeplot(sol, [[r, u(r)], [r, (D(u))(r)]], r = 0 .. 13.75, legend = [typeset("Curve: ", u(r)), typeset("Curve: ", (D(u))(r))]);

It's possible Bug in Ode Analyzer Assistant.

You must write equation like so:

y[1](x) -> y1(x)
y[2](x) -> y2(x)

diff(y1(x), x$2)+3*(diff(y1(x), x$1))+2*y1(x)+2*y2(x) = 3,

diff(y2(x), x$2)+diff(y2(x), x$1)+2*y2(x)+2*y1(x) = 8;

Regards Mariusz.

maybe like so:

expand(eval(pdsolve({diff(u(x, z, t), z)+C = 0, u(x, a, t) = 0}), a = h(x, t)));

#u(x, z, t) = C*h(x, t)-C*z

Maple has this function Totient(n) built-in.This command was introduced in Maple 2016. Using showstat function we can preview the code.

with(NumberTheory):
showstat(Totient);
showstat(PrimeFactors);

NumberTheory:-Totient := proc(n::{posint, And(algebraic,Not({boolean, `in`, complexcons, extended_numeric}))}, $)
local prime_factors, p;
   1   if not type(n,'posint') then
   2       return ('procname')(n)
       elif isprime(n) then
   3       return n-1
       else
   4       prime_factors := NumberTheory:-PrimeFactors(n);
   5       return n*mul(p-1,`in`(p,prime_factors))/convert(prime_factors,'`*`')
       end if
end proc

NumberTheory:-PrimeFactors := proc(n::{integer, And(algebraic,Not({boolean, `in`, complexcons, extended_numeric}))}, $)
local i;
   1   if n = 0 then
   2       error "cannot represent all prime factors of %1", n
       elif type(n,{'negint', 'posint'}) then
   3       return {seq(i[1],`in`(i,ifactors(n)[2]))}
       else
   4       return ('procname')(n)
       end if
end proc

Another code you can find on this webpage: http://oeis.org/A000010 see:MAPLE -> # version 2.

Have fun !

If yours integral is:

int(sqrt(ln(x)/x^2-1/x), x)

then Maple can't find it. Possible it has No closed form.

See: https://en.wikipedia.org/wiki/Nonelementary_integral.

Mathematica, Rubi,Axiom,SymPy, Maxima  cannot do it, either.

Second one:

int((ln(x)^(a-1)/x^2-1/x)^(1/2),x)

the same case.

EDITED:

If You want approximate integral by series see worksheet.

Approximate_indefine_integral_by_Series.mw

convert(exp(x), Sum, dummy = n)

or:

convert(exp(x), FormalPowerSeries)

For Simple function (2 methods):

you can use inverse ztransform:

Sum(x^n*invztrans(eval(exp(x), x = 1/x), x, n), n = 0 .. infinity)#only works if invztrans can find transfrom.

or n-th derivative:

Sum(x^n*(eval(diff(exp(x), x$n), x = 0))/factorial(n), n = 0 .. infinity)#only works if n-th derivative can find.

Only another way to calculate Julian date.

Formula from:http://aa.usno.navy.mil/faq/docs/JD_Formula.php

Compute the JD corresponding to 2018 January 11, 18h30m30s UT.

Substituting Y = 2018, M = 1, D = 11,h = 18, m = 30, s = 30;

restart:
JD := proc (Y, M, D, h, m, s) local x; 
x := evalf(367*Y-trunc((7/4)*Y+(7/4)*trunc((1/12)*M+3/4))+trunc((275/9)*M)+D+1721013.5-
(1/2)*signum(100*Y+M-190002.5)+1/2+(1/24)*h+(1/1440)*m+(1/86400)*s); end proc:

JD(2018, 1, 11, 18, 30, 30);

# 2.458130271*10^6

It may not possible to find anti-derivatives(closed form).

https://math.stackexchange.com/questions/1469123/integral-of-ex2

https://math.stackexchange.com/questions/168860/functions-cannot-be-integrated-as-simple-functions?rq=1

Mathematica gives No solution.

An aproximation only can be done e.g by series:

Download Only_aproximation.mw

Use Maple built function intsolve with method Neumann or with this  paper https://arxiv.org/ftp/arxiv/papers/1309/1309.6311.pdf see another attached file (Fredholm_integral_2ver.)

Download _Fredholm_integral.mw

_Fredholm_integral_2ver.mw

Only for k=1,4,5,6,7 is real solution.

Download sol.mw